Question

Find the area of the quadrilateral whose vertices taken in order, are $(-4,-2),(-3,-5),(3,-2)$ and $(2,3)$.

Answer 1

Let $A(–4,–2)$, $B(–3,–5)$, $C(3,–2)$ and $D(2,3)$ be the vertices of the quadrilateral $ABCD$.

Area of a quadrilateral $ABCD=$ Area of $△ABC$+ Area of $△ACD$

By using a formula for the area of a triangle $=\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_2)+x_3(y_1-y_2)|$

Area of $△$ABC

=$\frac{1}{2}[-4(-5+2)+-3(-2+2)+3(-2+5\left)\right]$

$=\frac{1}{2}[12+9]$

$=\frac{21}{2}$sq.units

Area of $△$ACD=$\frac{1}{2}[-4(3+2)+-2(-2+2)+3(-2-3\left)\right]$

$=\frac{1}{2}[-20-15]$

$=\frac{35}{2}$sq.units

$\therefore$Area of quadilateral=$\frac{21}{2}+\frac{35}{2}=\frac{56}{2}=28$ sq.unit